A particle moves along a straight line such that its displacement at any time t is given by

s = t³ - 6t² + 3t + 4 metres

The velocity when the acceleration is zero is

(a) 3 ms^-1

(b) -12 ms^-1

(c) 42 ms^-2

(d) -9 ms^-1

(d)
Velocity, v=  ds/dt=3t^2  -12 t+3
Acceleration a = ds/dt = 6t - 12; For a = 0 we have, 0 = 6 t - 12 or t= 2s. Hence, at t = 2 s the velocity will be
v = 3 × 22 – 12 × 2 + 3 = -9 ms-1

(d) -9m/s we know that differentiation of displacement gives velocity i.e. 3t^2-12t+3.now we know that differentiation of velocity gives acceleration i.e.6t-12but here acceleration is zero;therefore,6t-12=0;t=2secnow put the value of `t` in velocity we get -9m/sThank You

hello student

since ,Velocity, v= ds/dt=3t^2 -12 t+3
Acceleration a = ds/dt = 6t - 12; For a = 0 we have, 0 = 6 t - 12 or t= 2s. Hence, at t = 2 s the velocity will be
v = 3 × 2^2 – 12 × 2 + 3 = -9 ms-1

Hope it helps
Thankyou

Dear Student

Given
s= t³ − 6t² + 3t + 4
v = ds/dt​
So v = 3t² −12t + 3.............(1)
again differentiate, a = dv/dt​
6t − 12=0
t = 2s
Putting value of t in eq. (1)
v = 3(2)² −12×2 +3
v = −9ms⁻¹

I hope this answer will help you.
Thanks & Regards
Yash Chourasiya